What is the tangent of f(x)=4x(x^2-4x+5)^7,\at x=2 ?

The tangent of f(x)=4x(x^2-4x+5)^7,\at x=2 is y=4x

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tangent of f(x)=4x(x^2-4x+5)^7,\at x=2

tangent of

f (x) = 4 x (x^{2} - 4 x + 5)^{7}, at x = 2

y = 4 x

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

f (x) = 4 x (x^{2} - 4 x + 5)^{7} : \frac{df ( x )}{d x} = 4 ((x^{2} - 4 x + 5)^{7} + 7 x (x^{2} - 4 x + 5)^{6} (2 x - 4))

Compute the slope of

f (x) = 4 x (x^{2} - 4 x + 5)^{7}

x = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 8) : y = 4 x

y = 4 x

\int (x^{2} - 3)^{4} (2 x) d x

f (x) = e^{4 x}

\frac{s}{s ^{2} + 6 s + 25}

\frac{d}{d x} (\frac{3 - 2 x - x ^{2}}{x ^{2} - 6})

\frac{d}{d x} (x^{4} + 4 x^{2})

What is the tangent of f(x)=4x(x^2-4x+5)^7,\at x=2 ?
The tangent of f(x)=4x(x^2-4x+5)^7,\at x=2 is y=4x